Double indicator method, Chemistry tutorial

Introduction:

Specifically, only one kind of indicator was utilized for each of the cases and only one reactant was being quantitatively analyzed. Even in the last chapter where an indirect process of analysis was conducted that is Back titration, we utilized only one indicator since the reactants dictated the requirement for the employ of one indicator.

Though, there are cases where more than one indicator is needed for analyzing our model. We shall look into one of such reactions in this chapter as an instance of another technique available to you in volumetric analysis. This technique becomes useful whenever you have a mixture of substances making up a solution.

For example, if we have a solution containing a mixture of NaOH and Na2CO3 , that we wish to analyse, the utilize of one indicator to indicate the end of the reaction would be misleading since although an end point was indicated, the 2 substances couldn't have each reacted through the acid completely. To buttress this point, if HCl was titrated by this solution, the reactants that will take place in the same solution are:

1                NaOH + HCl         →     NaCl + H2O

11 a          Na2CO3   + HCl     →     NaHCO3 + HCl

11 b        NaNCO3   + HCl        →     NaCl +H2O + CO2

Assuming phenolphthalein is utilized as indicator, and then the pink colour of the indicator is discharged whenever reactions 1 and 11a are complete. This is actually the 1st step of the titration. At this point, all the NaOH is neutralized together by half of the Na2CO3 the other half of Na2CO3 will transform to NaHCO.

If methyl orange is now inserted and an additional quantity of acid is added, the amount of acid needed will be that necessary to complete reaction 11b. At this stage, the other half of the carbonate is neutralized. Since one mole of NaHCO3 is formal from one mole of Na2CO3 and therefore the quantity of HCl needed for reaction 11 a and 11 b will be similar.

Suppose the volume of the acid required reaching the end-pointing as indicated via the phenolphthalein is a cm3, and the volume of acid reacting with Na2CO3 is 2b then the volume of acid reacting by

NaOH is

                   (a + b) - 2bcm3   = (a - b) cm 

Where:

b cm3  = burette reading at methyl orange end point minus burette reading at phenolphthalein and a cm3   = burette reading at phenolphthalein end point.

Experiment:

Determination of the amount of NaOH and Na2CO3 in a specified mixture.

Requirements:

'A' is a standard solution of HCl of known molarity. 'B' is a solution of 2 bases containing unknown quantities of NaOH and Na2CO3.

Additional Materials

a) One Burette

b) Phenolphthalein and Methyl orange indicators

c) Unknown concentration of NaOH and Na2CO3 mixture.

Procedure:

1st step: Titration of mixture of bases against acid using Phenolphthalein.

1. Rinse the burette twice by solution 'A' containing HCl and fill it through similar acid. Note the initial burette reading.

2. Rinse the pipette through solution '13' containing a mixture of NaOH and Na2CO3 and take 25cm3 portion of this solution into a 250cm3 conical flask and add 2-3 drops of Phenolphthalein.

3. Add acid from the burette until pink colour disappears. Note the burette reading of the Phenolphthalein.

2nd Step: Continuation of titration through similar solutions, with no adding acid into the burette or base into the conical flask, using methyl orange (1 drop) as indicator.

4. Add one drop of methyl orange indicator to the colourless solution attained at the Phenolphthalein end point and continue adding HCl from the burette until one drop of acid gives colour. At this point the remaining NaHCO3 is neutralized. Note the burette reading again. This is the methyl orange end -point.

5. Now pipette another 25cm3 portion of solution 'B' (base) again into a clean conical flask, fill the burette through HCl solution and start the titration watchfully, 1st using Phenolphthalein. As in step one and then methyl orange as in step two.

6. Repeat the titration until 3 constant consequences are attained.

Problem:

From your average titre values, determine

a) The concentration in moles per dm3 of solution NaOH that has reacted

b) The concentration in moles per dm3 of solution Na2CO3 that has reacted.

c) The amount in grams of the NaOH in the mixture

d) The amount in mass of Na2CO3 in the mixture

e) The percent composition of NaOH in the mixture

f) The percent composition of Na2CO3 in the mixture Na = 23, 0 = 16. C = 12, H =1).

Results:

Burette Reading, cm3

   Volume of HCI cm3

Initial Reading

Phenolphthalein

  n endpoint

(a cm3)

 

Methyl orange

end

point

   Vol.   of  HCl      for

NaCO3

(bcm3)

HCl  for  the

Whole of

Na2 CO3

(2

bcm3 )

HCl

    For

NaO

H (a -

b)cm3)








Treatment of Results:

Calculations:

a) For the amount in mole,/ dm3 of NaOH:

NaOH+ HCl     →   NaCl+ H2O

1st. We find the amount in moles of HCl = Molar cone. X Volume /1000

= Molar conc. x (a - b ) cm3  /1000

= C moles

From the stoichiciometry. I mole HCl = 1 mole NaOH

Therefore Amount in moles of C moles HCl = C moles of NaOH

This C mole is enclosed in the 25cm3 of NaOH utilized in the titration.

Therefore 1000cm3 will enclose

= C moles x 1000 /25

= 40 C moles/ dm3   

b) For the amount in moles / dm3 of Na2CO3:

Na2CO3 + 2HCl   →    2NaCl + H2O

First, we find the amount in moles of HCI

= molar conc. x volume / 1000

= molar conc. x (2b) cm3 /1000

= D moles

From the stoichiciometry, 2 mole HCl = 1 mole NaOH

Therefore Amount in moles of D moles HCl = 1 /2 D moles of Na2CO3

Hence 1/2 D mole is contained in the 25cm3 of Na2CO3   used in the titration.

Therefore 1000cm3 will contain = 1/2 D moles x 1000 / 25

                                                  = 20 D moles/ dm3

c) Amount in g/dm3   of NaOH

Concentration in g/dm3 = molar mass x molar /conc. of NaOH

                                     = 40 x 40C

                                    = 1600C g/dm 

d) Amount in g/dm3 of Na2CO3=molar mass x molar /conc. of Na2CO3

                                          = 106 x 20D

                                          = 200D/gdm3  

e) Percentage of the NaOH in the mixture.

                 % of NaOH = Mass of NaOH perdm3 x 100 / the combined mass of the two bases.

f) Percentage of the NA2CO3 in the mixture

              % of Na2CO3   = Mass of N2CO3 perdm x 100 / The combined mass of the two bases

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